When solid lead(II) sulfide ore burns in oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas.

Question

When solid lead(II) sulfide ore burns in oxygen gas, the products are solid lead(II) oxide and sulfur dioxide gas.

How many grams of oxygen are required to react with 22.0 g of lead(II) sulfide?

How many grams of sulfur dioxide can be produced when 67.0 g of lead(II) sulfide reacts?

How many grams of lead(II) sulfide are used to produce 120. g of lead(II) oxide?

Answer 1

First thing is to write a correctly balanced equation:

2PbS(s) + 3O₂(g) ==> 2PbO(s) + 2SO₂(g) ... BALANCED EQUATION

To find g of O2 needed to react with 22.0 g of PbS, we must convert to moles and then back to grams.

moles PbS = 22.0 g x 1 mol PbS/239 g = 0.0921 moles

moles O2 needed = 0.0921 mol PbS x 3 mol O2/2 mol PbS = 0.138 mol O2 needed

grams O2 needed = 0.138 mol O2 x 32 g/mol = 4.42 g O2 needed

Use the same approach to find g of SO2 produced from 67.0 g PbS:

67.0 g PbS x 1 mol PbS/239 g x 2 mol SO2/2 mol PbS x 64 g SO2/mol SO2 = 17.9 g SO2 produced

Use the same approach to find grams of PbS used to produce 120. g of PbO:

120 g PbO x 1 mol PbO/223 g x 2 mol PbS/2 mol PbO x 239 g/mol PbS = 129 g PbS used